Colligative Properties - Depression of Freezing Point

Colligative Properties - Depression of Freezing Point

What is the freezing point depression of a solution containing 15 g of mannitol (molar mass = 182 g/mol) in 500 g of wat

Moles of mannitol = (15/182) ≈ 0.0824 mol . Molality = (0.0824/0.5) ≈ 0.1648 mol/kg . Δ Tf = 1.86 × 0.1648 ≈ 0.3065 K .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Colligative Properties - Depression of Freezing Point

A 0.2 molal Na₂SO₄ solution has a van’t Hoff factor of 2.5. What is the freezing point depression? ( Kf = 1.86 K kg mol⁻

Δ Tf = i · Kf · m . Δ Tf = 2.5 × 1.86 × 0.2 = 0.93 K .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Colligative Properties - Depression of Freezing Point

A solution is made by dissolving 20 g of a solute (molar mass = 40 g/mol) in water. If the freezing point depression is

Moles = (20/40) = 0.5 . Δ Tf = Kf · m . 0.465 = 1.86 × (0.5/w) , w = (0.5 × 1.86/0.465) ≈ 2 kg = 2000 g . Total mass = 20 + 2000 = 2020 g .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Colligative Properties - Depression of Freezing Point

What is the freezing point depression of a solution containing 12 g of a non-electrolyte solute (molar mass = 60 g/mol)

Moles of solute = (12/60) = 0.2 mol . Molality = (0.2/0.2) =appendicitis = 1 mol/kg . Δ Tf = Kf · m = 1.86 × 1 = 1.86 K .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Colligative Properties - Depression of Freezing Point

A 0.2 molal solution of NaBr in water has a freezing point depression of 0.744°C. What is the van’t Hoff factor? ( Kf =

Δ Tf = i · Kf · m . 0.744 = i × 1.86 × 0.2 . i = (0.744/1.86 × 0.2) = 2 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Colligative Properties - Depression of Freezing Point

A 0.2 molal solution of a solute in water has a freezing point depression of 0.744°C. If the solute dissociates into 2 i

Δ Tf = i · Kf · m . 0.744 = i × 1.86 × 0.2 . i = (0.744/1.86 × 0.2) = 2 . i = 1 + α (n - 1) , 2 = 1 + α (2 - 1) , α = 1 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Colligative Properties - Depression of Freezing Point

What is the molarity of a solution containing 9 g of glucose (molar mass = 180 g/mol) in 500 mL of solution?

Moles of glucose = (9/180) = 0.05 mol . Volume = 500 mL = 0.5 L. Molarity = (0.05/0.5) = 0.1 M .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Colligative Properties - Depression of Freezing Point

What is the freezing point depression of a solution containing 18 g of sucrose (molar mass = 342 g/mol) in 400 g of wate

Moles of sucrose = (18/342) ≈ 0.0526 mol . Molality = (0.0526/0.4) ≈ 0.1315 mol/kg . Δ Tf = 1.86 × 0.1315 ≈ 0.2446 K .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Colligative Properties - Depression of Freezing Point

A 0.1 molal solution of KCl has a van’t Hoff factor of 1.9. What is the freezing point depression? ( Kf = 1.86 K kg mol⁻

Δ Tf = i · Kf · m . Δ Tf = 1.9 × 1.86 × 0.1 ≈ 0.353 K .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Colligative Properties - Depression of Freezing Point

A 0.15 molal KCl solution has a van’t Hoff factor of 1.85. What is the freezing point depression? ( Kf = 1.86 K kg mol⁻¹

Δ Tf = i · Kf · m . Δ Tf = 1.85 × 1.86 × 0.15 ≈ 0.5162 K .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Colligative Properties - Depression of Freezing Point

What is the freezing point depression of a 0.2 molal solution of a non-electrolyte in water? ( Kf = 1.86 K kg mol⁻¹ )

Δ Tf = Kf · m . Δ Tf = 1.86 × 0.2 = 0.372 K .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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Colligative Properties - Depression of Freezing Point

A solution contains 7 g of a solute (molar mass = 140 g/mol) in 350 mL of solution. What is the molarity?

Moles of solute = (7/140) = 0.05 mol . Volume = 350 mL = 0.35 L. Molarity = (0.05/0.35) ≈ 0.1429 M .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

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